2025 AMC 10B Problem 4

1. Turn both numbers into ordinary numbers. In base seven, the left digit counts sevens, so $\underline{a}\,\underline{b}$ means $7a + b$. In base nine, the left digit counts nines, so $\underline{b}\,\underline{a}$ means $9b + a$.

2. Set them equal and simplify.
$$7a + b = 9b + a$$
Take $a$ and $b$ from both sides: $6a = 8b$. Divide by $2$:
$$3a = 4b.$$

1. Use what digits are allowed. Both $a$ and $b$ are digits of a base-seven number, so each is between $0$ and $6$. Each is also the first digit of a two-digit number, so neither is $0$.

1. Try the choices for $a$. Since $3a = 4b$, the number $3a$ must be a multiple of $4$. Check $a = 1, 2, \ldots, 6$: the values $3a$ are $3, 6, 9, 12, 15, 18$, and only $12$ is a multiple of $4$. So $a = 4$ and $b = 3$.

1. Check. $43$ in base seven is $7 \cdot 4 + 3 = 31$. $34$ in base nine is $9 \cdot 3 + 4 = 31$. They match.

So $a + b = 4 + 3 = 7$.

The answer is $\boxed{\textbf{(A)}\ 7}$.

A number written in another base is just a place-value sum, so a question about digits becomes an ordinary equation. One equation with two unknowns usually has many answers, and the extra fact that pins it down is the size limit on the digits. Always ask what values each digit is allowed to take.

Write each numeral in place-value form: $\underline{a}~\underline{b}$ in base seven is $7a + b$, and $\underline{b}~\underline{a}$ in base nine is $9b + a$. Setting them equal,
$$7a + b = 9b + a \quad\Longrightarrow\quad 6a = 8b \quad\Longrightarrow\quad 3a = 4b.$$

Now use the digit constraints. Both $a$ and $b$ appear as base-seven digits, so $0 \le a, b \le 6$, and both are leading digits of a two-digit numeral, so $a, b \ge 1$.

From $3a = 4b$ and $\gcd(3,4) = 1$, the factor $4$ must divide $a$. The only multiple of $4$ in the range $1 \le a \le 6$ is $a = 4$, which gives $b = 3$.

Check: $43_7 = 7 \cdot 4 + 3 = 31$ and $34_9 = 9 \cdot 3 + 4 = 31$. Hence $a + b = 7$.

The answer is $\boxed{\textbf{(A)}\ 7}$.